Ta có: \(A=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)\)
\(=\left(x+1\right)\left(x-6\right)\left(x-2\right)\left(x-3\right)\)
\(=\left(x^2-5x-6\right)\left(x^2-5x+6\right)\)
Đặt \(y=x^2-5x-6\)
nên ta có: \(A=y\left(y+12\right)\)
Ta có: A=y(y+12)(cmt)
\(=y^2+12y\)
\(=y^2+12y+36-36\)
\(=\left(y+6\right)^2-36\)
Ta có: \(\left(y+6\right)^2\ge0\forall y\)
\(\Rightarrow\left(y+6\right)^2-36\ge-36\forall y\)
Dấu '=' xảy ra khi \(\left(y+6\right)^2=0\)
\(\Leftrightarrow y+6=0\)
\(\Leftrightarrow x^2-5x-6+6=0\)
\(\Leftrightarrow x^2-5x=0\)
\(\Leftrightarrow x\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
Vậy: Giá trị nhỏ nhất của biểu thức \(A=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)\) là -36 khi x∈{0;5}
Ta có: \(A=\left[\left(x+1\right)\left(x-6\right)\right]\left[\left(x-2\right)\left(x-3\right)\right]\)
\(A=\left(x^2-5x-6\right)\left(x^2-5x+6\right)\)
\(A=\left(x^2-5x\right)^2-6^2\ge0-36=-36\)
\(\Rightarrow A\ge-36\)
Đẳng thức xảy ra khi và chỉ khi \(x^2-5x=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
Vậy \(A_{min}=-36\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
Ta có: \(A=\left(x+1\right).\left(x-2\right).\left(x-3\right).\left(x-6\right)\)
\(\Leftrightarrow A=\left[\left(x+1\right).\left(x-6\right)\right].\left[\left(x-2\right).\left(x-3\right)\right]\)
\(\Leftrightarrow A=\left(x^2-5x-6\right).\left(x^2-5x+6\right)\)
Đặt \(a=x^2-5x-6\)
Thay \(a=x^2-5x-6\) vào phương trình A, ta có:
\(A=a.\left(a+12\right)\)
\(\Leftrightarrow A=a^2+12a\)
\(\Leftrightarrow A=\left(a^2+12a+36\right)-36\)
\(\Leftrightarrow A=\left(a+6\right)^2-36\)
\(\Leftrightarrow A=\left(x^2-5x-6+6\right)^2-36\)
\(\Leftrightarrow A=\left(x^2-5x\right)^2-36\)
Vì \(\left(x^2-5x\right)^2\ge0\forall x\) \(\Leftrightarrow\) \(\left(x^2-5x\right)^2-36\ge-36\forall x\)
\(\Rightarrow A_{min}=-36\)
Dấu "=" xảy ra khi: \(\left(x^2-5x\right)^2=0\)
\(\Leftrightarrow x.\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=5\left(TM\right)\end{matrix}\right.\)
Vậy \(A_{min}=-36\) \(\Leftrightarrow\) \(\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
Ta có: \(A=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)\)
= [(x + 1)(x - 6)][(x - 2)(x - 3)]
\(=\left(x^2+x-6x-6\right)\left(x^2-2x-3x+6\right)\)
\(=\left(x^2-5x-6\right)\left(x^2-5x+6\right)\)
Đặt \(x^2-5x=a\), ta có:
A = \(\left(a-6\right)\left(a+6\right)=a^2-36\ge-36\forall a\)
Dấu = xảy ra khi:
\(a=0\Leftrightarrow x^2-5x=0\Leftrightarrow x\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
Vậy Min A = -36 khi x = 0 hoặc x = 5.
Ta có: A=(x+1).(x−2).(x−3).(x−6)A=(x+1).(x−2).(x−3).(x−6)
⇔A=[(x+1).(x−6)].[(x−2).(x−3)]⇔A=[(x+1).(x−6)].[(x−2).(x−3)]
⇔A=(x2−5x−6).(x2−5x+6)⇔A=(x2−5x−6).(x2−5x+6)
Đặt a=x2−5x−6a=x2−5x−6
Thay a=x2−5x−6a=x2−5x−6 vào phương trình A, ta có:
A=a.(a+12)A=a.(a+12)
⇔A=a2+12a⇔A=a2+12a
⇔A=(a2+12a+36)−36⇔A=(a2+12a+36)−36
⇔A=(a+6)2−36⇔A=(a+6)2−36
⇔A=(x2−5x−6+6)2−36⇔A=(x2−5x−6+6)2−36
⇔A=(x2−5x)2−36⇔A=(x2−5x)2−36
Vì (x2−5x)2≥0∀x(x2−5x)2≥0∀x ⇔⇔ (x2−5x)2−36≥−36∀x(x2−5x)2−36≥−36∀x
⇒Amin=−36⇒Amin=−36
Dấu "=" xảy ra khi: (x2−5x)2=0(x2−5x)2=0
⇔x.(x−5)=0⇔x.(x−5)=0
A = (x + 1)(x - 2)(x - 3)(x - 6) = [(x + 1)(x - 6)][(x + 2)(x + 3)] = (x^2 - 5x - 6)(x^2 - 5x +6)
Đặt x^2 - 5x = a
A = (a - 6)(a + 6) = a^2 - 36
Ta có: a^2 > 0 hoặc a^2 = 0 <=> A > -36 hoặc A = -36
Vậy giá trị nhỏ nhất của A là -36
